已知函数f(x)=sin(x π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数)
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已知函数f(x)=sin(x π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数)
若f(x)在区间[-π/3,0]上单调递增,且恰好能够取到f(x)的最小值2,试求a、b的值
若f(x)在区间[-π/3,0]上单调递增,且恰好能够取到f(x)的最小值2,试求a、b的值
![已知函数f(x)=sin(x π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数)](/uploads/image/z/5846008-40-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%28x+%CF%80%2F6%29+%2Bsin%28x-%CF%80%2F6%29+%2Bacosx+%2Bb%28a%2Cb%E2%88%88r%2C%E4%B8%94%E5%9D%87%E4%B8%BA%E5%B8%B8%E6%95%B0%EF%BC%89)
f(x)=sin(x π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数)
应该是f(x)=sin(x +π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数)?
f(x)=sin(x+ π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数)
=2sinxcosπ/6+acosx+b
=√3sinx+acosx+b
=√(3+a^2)sin(x+θ)+b (sinθ=a/√(3+a^2),cosθ=√3/√(3+a^2),
f(x)在区间[-π/3,0]上单调递增,且恰好能够取到f(x)的最小值2
∴-π/3+θ=-π/2,b-√(3+a^2)=2
θ=-π/6
a=-1,b=2+2=4
应该是f(x)=sin(x +π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数)?
f(x)=sin(x+ π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数)
=2sinxcosπ/6+acosx+b
=√3sinx+acosx+b
=√(3+a^2)sin(x+θ)+b (sinθ=a/√(3+a^2),cosθ=√3/√(3+a^2),
f(x)在区间[-π/3,0]上单调递增,且恰好能够取到f(x)的最小值2
∴-π/3+θ=-π/2,b-√(3+a^2)=2
θ=-π/6
a=-1,b=2+2=4
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